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Math Help - What did I did wrong in this U-sub Arc-Sec problem?

  1. #1
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    What did I did wrong in this U-sub Arc-Sec problem?

    (cos 3x) / sin3x * (√sinē(3x) -2)


    incase you aren't familiar with it, the formula for arc sec is

    1/a arcsec u/a + c

    a = -√2
    u = sin 3x
    du = 3 cos 3x

    so 1/3 du = cos 3x dx

    so I cancel out the top

    and make since it's in the proper format I make it

    I do 1/3 * a to get -√2/3 arcsec sin3x / -√2 + c

    what did I do wrong, if anything?
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  2. #2
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    Re: What did I did wrong in this U-sub Arc-Sec problem?

    I suggest that you use the u-sub. first; then, use the trig sub. second. Then resub back in terms of x. You are correct, you divide the 3 both side.
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    Last edited by Cbarker1; December 13th 2012 at 08:56 PM.
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  3. #3
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    Re: What did I did wrong in this U-sub Arc-Sec problem?

    Quote Originally Posted by Cbarker1 View Post
    I suggest that you use the u-sub. first; then, use the trig sub. second. Then resub back in terms of x.
    I appreciate the response but I'm very confused.

    How did you derive sin(3x) to get 1/3 cos 3x?

    sorry, maybe I'm being really stupid but I thought it'd be 3cos(3x)
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  4. #4
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    Re: What did I did wrong in this U-sub Arc-Sec problem?

    Quote Originally Posted by Cbarker1 View Post
    I suggest that you use the u-sub. first; then, use the trig sub. second. Then resub back in terms of x.
    Don't stress too much btw, I'll see the answer key tommorow but it was just bothering me I couldn't get it
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